Question

The average of $m\sin \left(m^{\circ }\right)$ where $m=2,4,6,\cdots ,180$ is equal to
(correct answer + 1, wrong answer - 0.25)

• A. $\cot {89}^{\circ }$
• B. $\cot 1^{\circ }$
• C. $\tan 1^{\circ }$
• D. $90\sin {89}^{\circ }$

Answer 1

The correct option is B $\cot 1^{\circ }$

Let $\frac{2\sin 2^{\circ }+4\sin 4^{\circ }+6\sin 6^{\circ }+\cdots +180\sin {180}^{\circ }}{90}=x$

Multiplying both sides by $\sin 1^{\circ },$ we get
$2\sin 2^{\circ }\sin 1^{\circ }+2\left(2\sin 4^{\circ }\sin 1^{\circ }\right)+\cdots +89\left(2\sin {178}^{\circ }\sin 1^{\circ }\right)=90x\sin 1^{\circ }$

Using the identity, $2\sin A\cos B=\cos (A-B)-\cos (A+B)$ we get
$(\cos 1^{\circ }-\cos 3^{\circ })+2(\cos 3^{\circ }-\cos 5^{\circ })+\cdots +89(\cos {177}^{\circ }-\cos {179}^{\circ })=90x\sin 1^{\circ }$
$\Rightarrow \cos 1^{\circ }+\cos 3^{\circ }+\cos 5^{\circ }+\cdots +\cos {175}^{\circ }+\cos {177}^{\circ }-89\cos {179}^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow \cos 1^{\circ }+(\cos 3^{\circ }+\cos {177}^{\circ })+\cdots +(\cos {89}^{\circ }+\cos {91}^{\circ })-89\cos {179}^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow \cos 1^{\circ }+0+\cdots +0+89\cos 1^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow 90\cos 1^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow x=\cot 1^{\circ }$

Alternate Solution:
$S=2\sin 2^{\circ }+4\sin 4^{\circ }+\cdots 178\sin {178}^{\circ }+180\sin {180}^{\circ }$
$S=2[\sin 2^{\circ }+2\sin 4^{\circ }+\cdots +89\sin {178}^{\circ }]\cdots \left(1\right)$

$S=2[89\sin {178}^{\circ }+88\sin {176}^{\circ }+\cdots +\sin 2^{\circ }]\cdots \left(2\right)$

Adding $\left(1\right)+\left(2\right),$ we get
$2S=2\left[90\right(\sin 2^{\circ }+\sin 4^{\circ }+\cdots +\sin {178}^{\circ }\left)\right]$
$S=90(\sin 2^{\circ }+\sin 4^{\circ }+\cdots +\sin {178}^{\circ })$

By using sine summation series,
$S=90\frac{\sin \left(\frac{n\theta }{2}\right)}{\sin \left(\frac{\theta }{2}\right)}\sin \left(\frac{(n+1)\theta }{2}\right)$

put $n=89,\theta =2^{\circ }$
$S=90\frac{\sin {89}^{\circ }}{\sin 1^{\circ }}\sin {90}^{\circ }$
$S=90\frac{\cos 1^{\circ }}{\sin 1^{\circ }}=90\cot 1^{\circ }$

Average value $=\frac{90\cot 1^{\circ }}{90}=\cot 1^{\circ }$